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In one of the proofs in my notes the Fejér Kernel ($K_{n}$ below) is plucked out of an absolute value seemingly for free. On the previous page it is remarked that this function is always non-negative. I can't see why this would be automatic. Why is this true?

We have $K_{n} = \sum\limits_{j=-n}^{n} D_{j}(t)$, where $D_{n} = \sum\limits_{j=-n}^{n}e^{ijt}$.

It is not automatic that $e^{ijt}$ is non-negative (or even real) for every value of $t\in [-\pi,\pi)$, $j\in\mathbb{Z}$.

This is what confuses me.

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@Srivatsan Narayanan: You mean $K_{n-1}$. –  AD. Oct 2 '11 at 20:45
    
@AD Yes, thanks for catching it. Actually, now that you pointed it out, there seem to be multiple issues in my formula, apart from $K_{n-1}$. The $1/n$ shouldn't appear. The question has index $j$ running from $-n$ to $n$, whereas the wikipedia has the corresponding index from $0$ to $n-1$. The two definitions of Fejer kernel seem to be quite different. I'll just remove my comment, and put up a new one with just the link. –  Srivatsan Oct 2 '11 at 20:54
    
See the wikipedia page. It has a closed-form expression that makes it evident that the Fejer kernel is always nonnegative wherever it is defined. (Note that wikipedia defines it in a somewhat different way though. In fact, my previous comment contained a formula that was just read off the wikipedia page; I removed it since it seems to have many errors.) –  Srivatsan Oct 2 '11 at 20:57
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Errors or not, what you pointed out drew my attention to the formula involving fractions of sin() functions squared in my notes which clarified the issue. So thank you. –  Kyle Schlitt Oct 2 '11 at 22:58
    
@Srivatsan Narayanan: I see what you mean! I am used to think of $K_n$ as $K_n(t)=\frac{1}{n+1}\sum_0^n D_k(t)$, where $D_k$ is the Dirichlet kernel and $n\ge0$ (is an integer). –  AD. Oct 3 '11 at 12:31

3 Answers 3

up vote 3 down vote accepted

Using the formula for the partial sum of a geometric series, the Dirichlet kernel is $$ D_n(t)=\frac{e^{i(2n+1)t/2}-e^{-i(2n+1)t/2}}{e^{it/2}-e^{-it/2}}\tag{1}=\frac{\sin((2n+1)t/2)}{\sin(t/2)} $$ We will sum the middle form, again using the formula for the partial sum of a geometric series, to get $$\begin{align} F_N(t) &=\frac{1}{N}\sum_{n=0}^{N-1}D_n(t)\\ &=\frac{1}{N}\frac{e^{iNt}-2+e^{-iNt}}{(e^{it/2}-e^{-it/2})^2}\\ &=\frac{1}{2N}\frac{1-\cos(Nt)}{\sin^2(t/2)}\\ &=\frac{1}{N}\frac{\sin^2(Nt/2)}{\sin^2(t/2)}\tag{2} \end{align} $$

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First, notice that $$ D_n(t) = \sum_{k=-n}^n \exp(i k t) = \exp(-i t n) \sum_{k=0}^{2n} \exp(i k t) = \mathrm{e}^{-i t n} \frac{1-\exp(i t (2n+1))}{1-\exp(i t)} = \frac{\sin ( n t + \frac{t}{2})}{\sin(\frac{t}{2})} $$ Then by definition $K_{n} = \frac{1}{n} \sum_{k=0}^{n-1} D_k(t)$ we have: $$ \begin{eqnarray} K_n &=& \frac{1}{n \sin(t/2)} \sum_{k=0}^{n-1} \sin\left( k t+\frac{t}{2}\right) \\ &=& \frac{1}{n \sin(t/2)^2} \frac{1}{2}\sum_{k=0}^{n-1} \left( \cos(k t) - \cos((k+1)t)\right) \\ &=& \frac{1}{2 n \sin(t/2)^2} \left( 1 - \cos( n t) \right) = \frac{1}{n} \left( \frac{\sin(n t/2)}{\sin(t/2)} \right)^2 \ge 0 \end{eqnarray} $$

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I believe that the numerator in your final answer should be $\sin(nt/2)$. –  robjohn Oct 2 '11 at 22:08
    
@robjohn Thanks for the catch. I have corrected that now. –  Sasha Oct 2 '11 at 23:44

You can prove that

$$\frac{1}{n}\sum_{j=0}^{n-1} D_j(t)=\begin{cases} \frac{1-\cos{nt}}{(1-\cos{t})n} &\mbox{if } t\not=2k\pi \\ n & \mbox{if } t=2k\pi \end{cases}\ge0$$ by induction on $n$ if $t=2k\pi$ and by multiplying $D_j(t)$ by $e^{it}-1$ otherwise (a telescopic series will appear).

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multiplicating is not an English word –  GEdgar Oct 2 '11 at 20:32

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